The sum is 0.\overline{8},
which is rational. So the sum of two irrationals may be rational. That
should not be surprising; 2 -\sqrt{2}
and \sqrt{2}
are irrational, but their sum is the rational 2.
Problem 16
The sum is 0.262662666\mathop{\mathop{…}},
which is not repeating or terminating and thus is irrational. So the sum of
two irrationals, like \sqrt{2}
and \sqrt{2},
can be irrational, like 2\sqrt{2}.
P = 2π\sqrt{{5.1\over
32}} ≈ 2.5.
Note that 32 is an approximation to gravitational acceleration.
Problem 81
Approximating:
1.{1}^{10}
1.0{1}^{100}
1.00{1}^{1000}
1.000{1}^{10000}
1.0000{1}^{100000}
2.5937
2.7048
2.7169
2.7181
2.7183
0.95418e
0.99505e
0.99950e
0.99095e
1.00000e
The constant e ={\mathop{ lim}}_{n→∞}{(1 + {1\over
n})}^{n},
and the experimental results above bear this out. The first few digits converge
very quickly.
Round each of the following to the nearest tenth (one place after the
decimal) using round to nearest even, round to zero (truncation),
and round half-up:
Compute the following quantities with a computer or a calculator.
Write what type of computer/calculator you used andthe software package if it’s a computer. Compute it as
shown. Do not simplify the expression before computing it, and
do not re-enter the intermediate results into the calculator or
computer program. Also compute the expressions that do not include
1{0}^{16}
by hand exactly. There should be a difference between the exact result
and the displayed result in some of these cases. Remember to work from
the innermost parentheses outward.
(\kern 1.66702pt (\overbrace{0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1}10\text{ times})\kern 1.66702pt - 1) × 1{0}^{16},
where 1{0}^{16}
often is entered as 1\text{e}16.
If the result overflows (signals an error) on various calculators,
replace 1{0}^{16}
by 1{0}^{8}
in this and later portions.
The object of this first part is to demonstrate round-off error. The first to problems,
adding 0.1
repeatedly, may see no error if the device calculates in decimal. The
latter four parts should see some error regardless of the base
used.
Using Octave on a 64-bit Intel-based machine with the “short” display format:
- 1.1102 ⋅ 1{0}^{-16}
- 1.1102
0 : Sometimes errors cancel themselves out. Not every computational error
is bad.